interger overflow.

suppose all no have gcd 1

But a[i] <= 10^4 and m <= 10^4 so max(lcm) = 10^8 which is within bounds of int and long long.

can u explain strip in a tree please

no

try to find the lcm prime numbers example (3,5,7,11) it will be the multiplication of all

and we have a lot of primes under 10^4 so it can easily exceed long long

Oh, I was too dumb. (10^4)^(10^4) would be the upper bound. Thank you for replying, by the way!!!

have a look at this

please take a look at this

# Oops! That page doesn’t exist or is private.

#include

using namespace std;

long long int gcd(long long int a,long long int b)

{

if (b == 0)

{

return a;

}

return gcd(b, a % b);

}

long long int lcm(long long int a[],long long int n)

{

long long int ans = a[0];

```
for (long long int i = 1; i < n; i++)
{
ans = (a[i] * ans)/ (gcd(a[i], ans));
}
return ans;
```

}

int main() {

// your code goes here

int t;

cin>>t;

while(t–)

{

long long int n,m;

cin>>n>>m;

```
long long int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
long long int ans=lcm(a,n);
long long int max=ans;
long long int num=1;
for(long long int i=2;i<=m;i++)
{
long long int cur=(i*ans)/(gcd(i,ans));
if(cur>max)
{
max=cur;
num=i;
}
}
cout<<num<<endl;
}
return 0;
```

}

can you help to find the error in this?

LCM overflows range of int. Store prime factorization instead.

have a look here

look at this

here take a look at this please

beautiful approach,i must say!

Thank you, till now i got the point but now how will you find the smallest no… I guess your approach should be traverse from 1 to m and for each no. find the prime factors and multiply with original lcm ???